Hi to u too! Are you a boy or a girl?

Tha's way easy. 5+7=12

Hi. I'm a 12 yr old girl answering. There are two answers for this. Either 0.7 or 7/10.

I am doing very fine thank u. I'm a 12 yr old girl responding to you. How's math?

The answer is 19

f**k you

he was called simon

because it is

CAN U HELP ME

PLZZZZZZZ

A number divided by 25 leaves remainder 9 . what is that number?

\(x\equiv 9 \pmod {25} \\ \text{or}\\ x-9 = z\cdot 25\\ x = 9+z\cdot 25 \qquad z \in Z\\\\ \mathbf{ x = \dots,-41, -16, 9, 34, 59, 84,\dots }\)

Thanks this makes sense

x / 25 = 9 -> x = 225

Possibly like:

Hallo guest,

Why not?

Like so: 

\(\delta x=1.8-2=-0.2 \\\\ \delta y=x_1^2-x_2^2+2(x_1-x_2)=(x_1-x_2)(x_1+x_2)+2(x_1-x_2)\\=\delta x(x_1+x_2)+2\delta x=\delta x(x_1+x_2+2)=-0.2(1.8+2+2)=-1.16 \)

The answer is x = 0 y = 1 z = 0

How do i solve, cos(2x)=-cos(x)

\(\small{ \begin{array}{lrcl} & \cos{(2x)} &=& -\cos{(x)} \\ & \mathbf{\cos{(2x)} + \cos{(x)} } & \mathbf{=} & \mathbf{0} \\ \\ \hline \\ \text{Formula derivation } & \cos{(ux+vx)} = \cos{(ux)}\cos{(vx)} - \sin{(ux)}\sin{(vx)}\\ & \cos{(ux-vx)} = \cos{(ux)}\cos{(vx)} + \sin{(ux)}\sin{(vx)}\\ & \cos{(ux+vx)} + \cos{(ux-vx)} = 2 \cos{(ux)}\cos{(vx)} \\ & & ux+vx = nx &\qquad uv-vx = mx\\ & &(1)~~ u+v = n & \qquad (2)~~u-v = m\\ (1) + (2) & n+ m = 2u \qquad u=\frac{n+m}{2}\\ (1) - (2) & n- m = 2v \qquad v=\frac{n-m}{2}\\ \end{array} }\\ \small{ \boxed{~ \cos{(nx)} + \cos{(mx)} = 2 \cos{ \left(\frac{n+m}{2}x \right)}\cos{ \left(\frac{n-m}{2}x \right)} ~} } \)

\(n=2 \qquad m=1\\ u=\frac{2+1}{2} = \frac32\\ v= \frac{2-1}{2}=\frac12\\ \)

\(\small{ \begin{array}{rcl} \cos{(2x)} + \cos{(x)} &=& 2\cos{ \left(\frac32 x \right) }\cos{ \left(\frac12 x \right) } &=& 0\\ 2\cos{ \left(\frac32 x \right) }\cos{ \left(\frac12 x \right) } &=& 0 \qquad | \qquad : 2 \\ \cos{ \left(\frac32 x \right) }\cos{ \left(\frac12 x \right) } &=& 0 \qquad \text{ setting each factor to 0}\\ \end{array} }\\ \small{ \begin{array}{lrcll} \\ 1. & \cos{ \left(\frac32 x \right) } &=& 0 \qquad & | \qquad \pm\arccos{()} \\ & \frac32 x &=& \pm\arccos{(0)} \pm 2k\pi \qquad & | \qquad \arccos{(0)} = \frac{\pi}{2} \\ & x &=& \frac{2}{3} ( \pm\frac{\pi}{2} \pm 2k\pi )\\ &\mathbf{ x } & \mathbf {=} & \mathbf{ \pm\frac{\pi}{3} \pm \frac{4}{3}k\pi }\qquad & | \qquad k \in N \\ \end{array} }\\ \small{ \begin{array}{lrcll} \\ 2. & \cos{ \left(\frac12 x \right) } &=& 0 \qquad & | \qquad \pm\arccos{()} \\ & \frac12 x &=& \pm\arccos{(0)} \pm 2k\pi \qquad & | \qquad \arccos{(0)} = \frac{\pi}{2} \\ & x &=& 2 \cdot ( \pm\frac{\pi}{2} \pm 2k\pi )\\ &\mathbf{ x } & \mathbf {=} & \mathbf{ \pm\pi \pm 4\cdot k\pi }\qquad & | \qquad k \in N \\ \end{array} }\\\)

Press the 2nd button (top left) then use asin, acos, atan  (they mean the same as sin^-1, cos^-1 and tan^-1)

x^2 + 2x - 48 → (x + 8)(x - 6)

Assuming n is Newtons (usually N is used) then mass = Force/g where g ≈ 9.8 m/s²

So mass = 1323/9.8 = 135 kg

-a² is taken to mean -(a²) = -a² by most pieces of technical software because raising to a power takes precedence over negation. 

Not all technical software adopts this convention. Microsoft Excel, for example, takes the opposite view and sets -a² = a²

(which means that it calculates -1² = 1, but also calculates 0-1² = -1)

You should check what convention is adopted by whatever calculator/software you are using.

It depends on what it is a percentage of.

As a percentage of 7 it's 100%, as a percentage of 14 it's 50%, as a percentage of 1 it's 700% etc..

Use the fact that √60 = √(15*2^2) = 2√15  and the fact that a*a^7 = a^1*a^7 = a^(1+7) = a^8

This simplifies as follows:  \(-3(3z-1)-2(5z-9)\rightarrow -9z+3-10z+18\rightarrow -19z+21\)

If this is set equal to zero then z = 21/19 as Guest has indicated above.